Part b: setting up integrals
a circular ring is made of a material with uniform density p (kg/m²).
the ring has rectangular cross section, as shown in the figure to the
right. the outer radius of the ring is r.
we would like to calculate the total mass of the ring.
first we will use the approximation that the width of the ring, a, is very much smaller than
the outer radius of the ring r(a < < r).
a) on the figure to the right, label a small "chunk"
of the ring that subtends an arc de.
b) what is the mass of this chunk (dm) in terms of
p. a, b, r, and de?
c) express the mass of the entire ring as a single integral in the angle 0. (don't forget limits)
d) what is the total mass of this ring?

b) dm = ρab R dθ

c) m = ∫ ρab R dθ, from θ = 0 to θ = 2π

d) m = 2πρabR

Explanation:

b) We want to find the mass dm of a small chunk of the ring.

Mass is density times volume:

dm = ρ dV

Since a << R, we can approximate the volume as a rolled rectangular prism.  Therefore, the volume of the chunk is the area of the cross section times the arc length.

dV = ab R dθ

dm = ρab R dθ

c) The mass of the entire ring is the sum of the masses of all the chunks.

m = ∫ dm

m = ∫ ρab R dθ, from θ = 0 to θ = 2π

d) ρ, a, b, and R are constants, so:

m = ρabR ∫ dθ

Evaluating the integral:

m = ρabR (θ|0 to 2π)

m = ρabR (2π − 0)

m = 2πρabR

the independent variable, the one that is intentionally manipulated, is whether or not the car hits the barrier.

the dependent variable, the one that you measure the response in, is the distance that the washers travel.

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.